EasyMathematicsHigh School (11–12)Algebra › Matrices and DeterminantsNumericalBoard Exam

Let A = 2, -1; 3, 0 and B = 1, 4; -2, 5…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Let $A =

\displaystyle \begin{bmatrix} 2 & -1 \\ 3 & 0 \end{bmatrix}

$and$ B =

\displaystyle \begin{bmatrix} 1 & 4 \\ -2 & 5 \end{bmatrix}

$. Find the entry in row$ 2 $, column$ 2 $of$ AB $, and the entry in row$ 1 $, column$ 1 $of$ BA$.

Need a reminder? The key ideaShow

$(AB)_{ij}$ is row $i$ of $A$ multiplied by column $j$ of $B$ , and $AB$ exists when the columns of $A$ equal the rows of $B$ . Matrix multiplication is associative and distributive but not commutative. The entry of $N^k$ counts routes with $k$ steps.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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